📝 Natural domain of a function (14 MCQs)

📖 Explanation: The denominator vanishes when $x-2=0$ or $x+5=0$, i.e., at $x=2$ and $x=-5$. Since division by zero is undefined, those two points must be excluded while every other real number is allowed, giving the domain $(-\infty, -5)\cup(-5,2)\cup(2,\infty)$.

📖 Explanation: Factor the radican$x^{2}-4x+3=(x-1)(x-3)$. The product is non‑negative when $x\le1$ or $x\ge3$. Including the zeros gives the closed intervals at 1 and 3, so the domain is $(-\infty,1]\cup[3,\infty)$.

📖 Explanation: Both $\tan x$ and $\sec x$ are undefined wherever $\cos x=0$. This occurs at odd integer multiples of $\pi/2$. Consequently, each function shares the exact same set of permissible $x$-values, namely $\mathbb{R}\setminus\{(2k+1)\pi/2\mid k\in\mathbb{Z}\}$.

📖 Explanation: The natural domain consists of all input values $x$ for which the expression defining the function is mathematically meaningful and returns a real number. It is determined by the restrictions imposed by denominators, radicals, logarithms, and other operations that limit admissible inputs.

📖 Explanation: The denominator becomes zero when $x-3=0$, i.e., at $x=3$. No other restriction exists because the numerator can take any real value. Hence the function is defined for every real number except $x=3$, giving the domain $(-\infty,3)\cup(3,\infty)$.

📖 Explanation: The radicand requires $x-1\ge0$ → $x\ge1$. The denominator cannot be zero, so $x\neq2$. Combining these conditions yields the set $[1,2)\cup(2,\infty)$. The endpoint $x=1$ is permissible because the numerator becomes zero while the denominator stays non‑zero.

📖 Explanation: The expression under the square root must be strictly positive because it appears in the denominator. Solving $x^{2}-4>0$ gives $x<-2$ or $x>2$. Therefore the domain is the union of the two open intervals $(-\infty,-2)$ and $(2,\infty)$.

📖 Explanation: The argument of the logarithm must be positive. Factoring gives $(x-1)(x-3)>0$, which holds for $x<1$ or $x>3$. The endpoints where the product equals zero are excluded because $\ln 0$ is undefined. Hence the domain is $(-\infty,1)\cup(3,\infty)$.

📖 Explanation: The original fraction has a denominator that vanishes at $x=2$, creating a hole in its graph. After canceling the common factor, the simplified expression $x+2$ no longer shows that restriction, so it appears defined at $x=2$. Consequently, the domains differ: $g$ excludes $2$ whereas $h$ includes it.

📖 Explanation: The fraction must be non‑negative and the denominator cannot be zero. Solving $\frac{x-1}{x+2}\ge0$ yields $x\ge1$ (the zero at $x=1$ is allowed) and automatically excludes $x=-2$ because it lies outside the solution set. Hence the domain is $[1,\infty)$.

📖 Explanation: A cube root is defined for any real radicand, unlike even roots that require non‑negative arguments. Since $x^{2}-5x+6$ can take any real value, there is no restriction on $x$. Therefore the natural domain consists of all real numbers.

📖 Explanation: Both numerator and denominator must keep the quotient non‑negative and the denominator cannot be zero. Solving $\frac{(x-2)(x+2)}{x-3}\ge0$ gives intervals $[-2,2]$ and $(3,\infty)$. The point $x=3$ is excluded because it makes the denominator zero, while the zeros at $-2$ and $2$ are allowed, yielding the domain $[ -2,2]\cup(3,\infty)$.

📖 Explanation: Both square‑root terms require non‑negative arguments, giving $x\ge1$. The denominator $\sqrt{x}-\sqrt{x-1}$ never vanishes for $x\ge1$ because $\sqrt{x}>\sqrt{x-1}$. Hence the function is defined for all $x\ge1$ with no removable discontinuities, i.e., the domain is $[1,\infty)$.

📖 Explanation: The original radicand $\frac{x-4}{(x-3)(x+3)}$ is non‑negative for $(-3,3)$ and for $x\ge4$; $x=\pm3$ are excluded because they zero the denominator. Multiplying numerator and denominator by $x+3$ introduces an extra factor $(x+3)^2$ in the denominator, which still only forbids $x=-3$. No new restrictions appear, so the domain stays $(-3,3)\cup[4,\infty)$.